3.1 volt 400 mA laser driver

Question

I've done quite a bit of searching here and on Google, but I think I still need help.

Using either of these power supplies: 12V 1A or 5V 2A

How can I build a circuit to drive a laser diode with optimal specs of 3.1 V at 400 mA? I purchased via eBay, and the description specifically says ML101U29 laser diode. I found a data sheet for it. It is inside a casing, inside a heatsink.

I'm an electronics newbie and am still grappling with how electricity works and how to shape it in this manner. What I've learned so far is that the LM317 is a go-to part for power supplies/drivers. In fact, I even have an LM317-based circuit pre-assembled. I can adjust it to get ~3 volts out of it (measured via multimeter), but have no idea how to control or limit the mA to ensure my laser doesn't burn up, and unfortunately it does not provide detailed specs. In my testing, I was able to light the laser, but even after focusing, did not seem to generate much heat.

I also found this power supply kit from Adafruit that will output adjustable or 3.3 volts (very close to target). But, it let's through 1.25 A--way too much for my diode. My understanding is that I should be able to use a resistor to limit the 1.25 A to 400 mA, but I am uncertain if my understanding is accurate.

So, I'm reaching out to experts. What would you suggest? Thank you!

UPDATE: I ended up using a driver board I already had and just turning it up. I don't know what the true voltage going through the laser is, or the current, but it works. I turned it up very slowly to get to burning capability without frying the laser. I use an LM317 Adjustable Module pre-assembled, available from a few online retailers. The most interesting thing so far is that it requires a lot more current to burn light-colored wood. Using a marker to color it black results in a burn at a lower power, but is useless because it's black on black.

Answer 1

You've misread the data sheet. Nominal operating current (according to the data sheet) is ~200 mA. In fact, the 400 mA limit is mentioned nowhere in the data sheet, and I'd guess you got it from the eBay listing. A word of advice - when it comes to believing an eBay listing or a data sheet - go with the data sheet. With this in mind, your best bet (for consistent power levels) is

^{simulate this circuit – Schematic created using CircuitLab}

I recommend that you find a copy of the LM317 data sheet and read it. The circuit is one of the application circuits. The biggest drawback of the circuit is that the LM317 will dissipate nearly 2 watts, so you need a decent heat sink on it.

Going with the 5 volts supply/resistor is workable, but the resistor should be 10 ohms, in order to limit the current appropriately.

And don't forget to heatsink the laser.

Answer 2

Well, one thing about power supplies is they have to regulate either the voltage our the current, not both, at lease not at the same time. A 5V 2A power supply will supply up to 5V and up to 2A, as dictated by the load. Actually, some supplies do not regulate the current at all so much as just limit it to prevent excessive current draw the case of a short.

Diodes have an exponential I-V relationship - the current through a diode varies with the exponential of the voltage. This relationship is also temperature dependent. So ifyou have a diode, what you need to do is limit the current that it can draw from the supply. Doesn't matter if you have an LED or a laser diode; the fundamental principle of operation is the same and the electrical characteristics are the same. Also, the current directly determines the brightness, not the voltage.

Generally the datasheet will give a recommended current and forward voltage - in your case, 3.1V and 400mA. What you need to do is design a supply that will provide a constant 400 mA at around 3.1 volts. The simplest methods is to use a series current limiting resistor. If you have a 5V power supply, the voltage across the resistor at the operating point will be 5 - 3.1 = 1.9V. Now all you need to do is figure out how big of a resistor creates a 1.9 volt drop when 400 mA flows through it - 1.9 / 0.4 = 5 ohms. And you will need a rather beefy resistor as it will have to dissipate 0.4 * 1.9 = 1W of heat.

There are also ways of building constant current sources with an LM317 that would be immune to supply voltage variations.